Question: What is the extraneous solution to these equations? $\dfrac{x^2}{x + 7} = \dfrac{-3x + 28}{x + 7}$
Solution: Multiply both sides by $x + 7$ $ \dfrac{x^2}{x + 7} (x + 7) = \dfrac{-3x + 28}{x + 7} (x + 7)$ $ x^2 = -3x + 28$ Subtract $-3x + 28$ from both sides: $ x^2 - (-3x + 28) = -3x + 28 - (-3x + 28)$ $ x^2 + 3x - 28 = 0$ Factor the expression: $ (x + 7)(x - 4) = 0$ Therefore $x = -7$ or $x = 4$ At $x = -7$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -7$, it is an extraneous solution.